“2020年GeCAA理论第2题-地平论”的版本间的差异
Jingsong Guo（讨论 | 贡献） |
Jingsong Guo（讨论 | 贡献） |
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| （未显示同一用户的8个中间版本） | |||
| 第28行： | 第28行： | ||
''Solution'' | ''Solution'' | ||
| − | + | Assume the surface area of one side of the flat Earth is $$A$$. Let the angle | |
between the Sun and the flat Earth’s center axis be $$θ$$, where $$θ$$ is initially $$0°$$. As | between the Sun and the flat Earth’s center axis be $$θ$$, where $$θ$$ is initially $$0°$$. As | ||
the Sun’s rays are parallel, the power delivered to the Earth by the Sun will be | the Sun’s rays are parallel, the power delivered to the Earth by the Sun will be | ||
| 第35行： | 第35行： | ||
At equilibrium, this is the energy radiated away via blackbody radiation, so the | At equilibrium, this is the energy radiated away via blackbody radiation, so the | ||
equilibrium temperature $$T$$ satisfies | equilibrium temperature $$T$$ satisfies | ||
| − | $$S_⊙Acos θ = σ(2A)T^4$$ 2.0 | + | $$S_⊙Acos θ = σ(2A)T^4$$ <div style="float:right">(''2.0 points'')</div> |
| + | |||
| + | |||
, where the factor of 2 comes from the fact that the flat Earth would radiate energy | , where the factor of 2 comes from the fact that the flat Earth would radiate energy | ||
from both sides. | from both sides. | ||
This yields | This yields | ||
| − | T(θ) = 4 | + | $$T(θ) = \sqrt[4]{\dfrac{ |
| − | + | S_⊙ cos θ}{ | |
| − | + | 2σ}}$$ | |
| − | 2σ | + | |
| − | and we wish to find the value of | + | and we wish to find the value of $$θ_1$$ such that $$T(θ_1) = T(0) − ΔT$$. |
Thus, | Thus, | ||
| − | 4 | + | |
| − | + | $$\sqrt[4]{\dfrac{S_⊙ cos θ_1} | |
| − | + | {2σ}} | |
| − | 2σ | + | = \sqrt[4]{{S_⊙} |
| − | = 4 | + | {2σ}} |
| − | + | − ΔT$$ <div style="float:right">(''2.0 points'')</div> | |
| − | + | ||
| − | 2σ | + | |
| − | − ΔT 2.0 | + | $$cos θ_1 = |
| − | cos | ||
| − | 1 − ΔT 4 | + | (1 − ΔT \sqrt[4]{\dfrac{ |
| − | + | 2σ} | |
| − | 2σ | + | {S⊙}} |
| − | S⊙ | + | )^4$$ |
| − | + | <div style="float:right">(''1.0 points'')</div> | |
| − | 1.0 | + | $$= |
| − | = | + | (1 − \sqrt[4]{\dfrac{ |
| − | + | 2 × 5.67 × 10^{−8}} | |
| − | 1 − 4 | + | {1366}} |
| − | + | )^4$$ | |
| − | 2 × 5.67 × | + | |
| − | 1366 | + | $$cos θ_1 = 0.9880$$ <div style="float:right">(''1.0 points'')</div> |
| − | + | ||
| − | cos | + | |
Now, we find the time it takes for the axis to make such an angle with the Sun. On | Now, we find the time it takes for the axis to make such an angle with the Sun. On | ||
the celestial sphere, let O be the center of precession, Z be the current direction of | the celestial sphere, let O be the center of precession, Z be the current direction of | ||
| − | the axis, and X be direction of the axis when it makes an angle of | + | the axis, and X be direction of the axis when it makes an angle of $$θ_1$$ with the sun, |
| − | + | ||
| − | so | + | so $$\angle ZCX = θ_1$$. If ϵ is the radius of precession, then$$ \angle OCZ = \angle OCX = ϵ$$. <div style="float:right">(''1.0 points'')</div> |
| + | |||
| + | |||
By the spherical Law of Cosines on angle O of spherical triangle OXZ, we have | By the spherical Law of Cosines on angle O of spherical triangle OXZ, we have | ||
| − | cos | + | |
| − | = | + | $$cos θ_1 = cos ϵ cos ϵ + sin ϵ sin ϵ cos(O)$$ |
| − | + | ||
| − | + | $$= cos^2 ϵ + sin^2 ϵ cos(O)$$ | |
| − | cos | + | |
| − | + | $$O = cos^{−1} | |
| − | + | ( | |
| − | 1.0 | + | \dfrac{cos θ_1 − cos^2 ϵ} |
| − | + | {sin^2 ϵ} | |
| − | 2π | + | )$$ |
| − | × P = | + | <div style="float:right">(''1.0 points'')</div> |
| − | P | + | |
| − | 2π | + | |
| − | + | $$\therefore Δt = \dfrac{\angle O} | |
| − | + | {2π} | |
| − | cos | + | × P $$ |
| − | + | ||
| − | + | $$= | |
| − | 1.0 | + | \dfrac{P} |
| − | = | + | {2π} |
| − | 25800 | + | ×cos^{−1} |
| − | 2π | + | (\dfrac{cos θ_1 − cos^2 ϵ} |
| − | × | + | {sin^2 ϵ} |
| − | + | )$$ | |
| − | 0.9880 − | + | <div style="float:right">(''1.0 points'')</div> |
| − | + | ||
| − | + | ||
| − | ≈ 1606 yr | + | $$= |
| − | Thus, the average temperature of the earth will go down by 1 °C in just over 1600 | + | \dfrac{25800} |
| − | years. 1.0 | + | {2π} |
| − | + | × cos^{−1} | |
| + | ( | ||
| + | \dfrac{0.9880 − cos^2 23.5°} | ||
| + | {sin^2 23.5°} | ||
| + | )$$$$≈ 1606 yr$$ | ||
| + | |||
| + | Thus, the average temperature of the earth will go down by 1 °C in just over 1600 years. <div style="float:right">(''1.0 points'')</div> | ||
2021年6月16日 (三) 22:50的最新版本
英文题目
2 Flat Earth (10 points)
A new model of the world is gaining in popularity among some people. These people believe in the “Flat Earth” view of the world, where the Earth is not a spheroid, but rather a circle with radius $$R_⊕$$. The central axis of the Earth (normal to the circle passing through its centre C) is passes through the observer’s zenith. This model must at least remain consistent with the observed phenomena, as listed below:
• The value of the solar constant is $$S_⊙ = 1366W/m^2$$.
• The Earth’s central axis precesses in a circle with a period 25800 years.
• The radius of the precession circle is 23.5°.
We assume that the Earth is a perfect blackbody radiator and the Sun is sufficiently far away that all sun rays are parallel. Let us also assume that the Sun’s current (initial) location is at the zenith.
Determine how many years it will take for the Earth’s equilibrium temperature to decrease by 1 °C.
中文翻译
官方解答
Solution
Assume the surface area of one side of the flat Earth is $$A$$. Let the angle between the Sun and the flat Earth’s center axis be $$θ$$, where $$θ$$ is initially $$0°$$. As the Sun’s rays are parallel, the power delivered to the Earth by the Sun will be $$S_⊙Acos θ $$at any given point in time.
At equilibrium, this is the energy radiated away via blackbody radiation, so the equilibrium temperature $$T$$ satisfies
$$S_⊙Acos θ = σ(2A)T^4$$
, where the factor of 2 comes from the fact that the flat Earth would radiate energy
from both sides.
This yields
$$T(θ) = \sqrt[4]{\dfrac{
S_⊙ cos θ}{
2σ}}$$
and we wish to find the value of $$θ_1$$ such that $$T(θ_1) = T(0) − ΔT$$. Thus,
$$\sqrt[4]{\dfrac{S_⊙ cos θ_1} {2σ}} = \sqrt[4]{{S_⊙} {2σ}} − ΔT$$
$$cos θ_1 =
(1 − ΔT \sqrt[4]{\dfrac{
2σ}
{S⊙}}
)^4$$
$$= (1 − \sqrt[4]{\dfrac{ 2 × 5.67 × 10^{−8}} {1366}} )^4$$
$$cos θ_1 = 0.9880$$
Now, we find the time it takes for the axis to make such an angle with the Sun. On
the celestial sphere, let O be the center of precession, Z be the current direction of
the axis, and X be direction of the axis when it makes an angle of $$θ_1$$ with the sun,
so $$\angle ZCX = θ_1$$. If ϵ is the radius of precession, then$$ \angle OCZ = \angle OCX = ϵ$$.
By the spherical Law of Cosines on angle O of spherical triangle OXZ, we have
$$cos θ_1 = cos ϵ cos ϵ + sin ϵ sin ϵ cos(O)$$
$$= cos^2 ϵ + sin^2 ϵ cos(O)$$
$$O = cos^{−1} ( \dfrac{cos θ_1 − cos^2 ϵ} {sin^2 ϵ} )$$
$$\therefore Δt = \dfrac{\angle O}
{2π}
× P $$
$$= \dfrac{P} {2π} ×cos^{−1} (\dfrac{cos θ_1 − cos^2 ϵ} {sin^2 ϵ} )$$
$$=
\dfrac{25800}
{2π}
× cos^{−1}
(
\dfrac{0.9880 − cos^2 23.5°}
{sin^2 23.5°}
)$$$$≈ 1606 yr$$
Thus, the average temperature of the earth will go down by 1 °C in just over 1600 years.
